关于python的多线程的Condition问题，制作了一个实验

实验代码如下，可以直接运行。实验结论可能跟文档的解释不太一样。

import threading
import time

condi = threading.Condition()
N =1
condi2 = threading.Condition()

def tar01():
    global N
    with condi:
        print('%s is started'%threading.current_thread().name)
        print(N)
        N +=1
        time.sleep(0.1)
        print('in %s,wait01 is triggered'%threading.current_thread().name)
        condi.wait()
        print('in %s,notify 01 is triggered'%threading.current_thread().name)
        condi.notify()
        print('in %s,notify 01 finished'%threading.current_thread().name)


def tar02():
    global N
    with condi:
        print('%s is started'%threading.current_thread().name)
        print(N)
        N +=1
        time.sleep(0.1)
        print('in %s,notify 02 is triggered'%threading.current_thread().name)
        condi.notify()
        print('in %s,wait 02 is triggered'%threading.current_thread().name)
        condi.wait()
        print('in %s,wait 02 finished'%threading.current_thread().name)


def tar03():
    global N
    with condi:
        print('%s is started'%threading.current_thread().name)
        print(N)
        N += 1
        time.sleep(0.1)
        print('in %s,notify 03 is triggered'%threading.current_thread().name)
        condi.notify()
        print('in %s,wait 03 is triggered'%threading.current_thread().name)
        condi.wait()
        print('in %s,wait 03 finished'%threading.current_thread().name)
        # print('in %s,notify 03 is triggered'%threading.current_thread().name)
        # condi.notify()
        # print('in %s,notify 03 finished'%threading.current_thread().name)


def tar04():
    with condi2:
        print('%s is started'%threading.current_thread().name)
        for i in range(1,7,2): #打印1到6之间的基数，因为起始点是1，步长是2
            print(i)
            time.sleep(0.1)
            print('in %s,notify 01 is triggered'%threading.current_thread().name)
            condi2.notify()
            print('in %s,wait 01 is triggered'%threading.current_thread().name)
            condi2.wait()
            print('in %s,wait 01 finished'%threading.current_thread().name)
            # print('in %s,notify 01 is triggered'%threading.current_thread().name)
            # condi2.notify()
            # print('in %s,notify 01 finished'%threading.current_thread().name)


def tar05():
    with condi2:
        print('%s is started'%threading.current_thread().name)
        for i in range(2,7,2): #打印1到6之间的偶数，因为起始点是2，步长是2
            print(i)
            time.sleep(0.1)
            print('in %s,notify 02 is triggered'%threading.current_thread().name)
            condi2.notify()
            print('in %s,wait 02 is triggered'%threading.current_thread().name)
            condi2.wait()
            print('in %s,wait 02 finished'%threading.current_thread().name)

key = input('press Enter to go mod A,else: press other key:  ')

if key == '':
    for i in range(2):
        # 这个for循环的含义就是每轮开启3个线程，由于第1轮循环开启的时候，第1轮循环开启的线程尚未结束，
        # 所以第2轮同时又开启了3个循环。并且，由于第二轮开启的时候，第一轮循环里的第一个受制于condition的函数尚未结束，
        # 并且这6个线程在代码中还受制于同一个condition，所以他们在实际运行中也是受制于同一个condition的
        # 所以，他们必然要依次开启一同结束。受到同一个condition制约的线程是同时结束的
        threading.Thread(target=tar01).start()
        threading.Thread(target=tar02).start()
        threading.Thread(target=tar03).start()
else:
    # t1 = threading.Thread(target= tar01) # 这三次个线程即使没有被使用，也被创建了出来，并占用了头两个线程的编号
    # t2 = threading.Thread(target= tar02)
    # t3 = threading.Thread(target= tar03)

    threading.Thread(target=tar04).start()
    threading.Thread(target=tar05).start()

# print('+++++++++++++++finished+++++++++++++++')

# 运行结果证明：
# 1. 在循环执行两次两个tar01和tar02时，tar02唤醒的原因是tar02.wait()，
# 2. notify()函数是用来告诉Condition，如果再没有新的线程需要等待的，那么本线程可以结束了的，
#    以及如果线程池里只剩下一个处于等待状态的线程时，这个线程也被**并运行这个线程剩余的代码
#下面是不同情况的具体总结：

# +++++++++++++++++++
# 对于顺序执行tar01 tar02 tart03的情况
if 1 == 1:
    pass
    # press Enter to go mod A,else: press other key:
    # Thread-1 is started
    # 1
    # in Thread-1,wait01 is going to work
    # Thread-2 is started
    # 2
    # in Thread-2,notify 02 is going to work
    # in Thread-2,wait 02 is going to  work
    # Thread-3 is started
    # 3
    # in Thread-3,notify 03 is going to work
    # in Thread-3,wait 03 is going to  work
    # Thread-4 is started
    # 4
    # in Thread-4,wait01 is going to work
    # Thread-5 is started
    # 5
    # in Thread-5,notify 02 is going to work
    # in Thread-5,wait 02 is going to  work
    # Thread-6 is started
    # 6
    # in Thread-6,notify 03 is going to work
    # in Thread-6,wait 03 is going to  work
    # in Thread-1,notify 01 is going to  work
    # in Thread-1,notify 01 finished
    # in Thread-5,wait 02 finished
    # in Thread-2,wait 02 finished
    # in Thread-3,wait 03 finished
    # in Thread-4,notify 01 is going to  work
    # in Thread-4,notify 01 finished
    # in Thread-6,wait 03 finished
    #
    # Process finished with exit code 0
# +++++++++++++++++++++++++++++++++++++++++
#对于tar04和tar05的情况：假如每个函数循环的总次数是偶数，
if 1 == 1:
    pass
    # 比如：
    # tar04执行for循环的次数为3次，for i range(1,7,2) 循环打印的是1，3，5
    # tar05执行for循环的次数为3次，for i range(2,7,2) 循环打印的是2，4，6
    # 那么tar04写成先wait再notify，tar05写成先notify再wait的话，这样的两个线程是可以都结束的。
    # 程序运行结果如下：
    # press Enter to go mod A,else: press other key:  p
    # Thread-1 is started
    # 1
    # in Thread-1,wait 01 is triggered #这两部清晰的证明wait会导致下一个Condition函数被release
    # Thread-2 is started
    # 2
    # in Thread-2,notify 02 is triggered
    # in Thread-2,wait 02 is triggered
    # in Thread-1,wait 01 finished
    # in Thread-1,notify 01 is triggered
    # in Thread-1,notify 01 finished
    # 3
    # in Thread-1,wait 01 is triggered
    # in Thread-2,wait 02 finished
    # 4
    # in Thread-2,notify 02 is triggered
    # in Thread-2,wait 02 is triggered
    # in Thread-1,wait 01 finished
    # in Thread-1,notify 01 is triggered
    # in Thread-1,notify 01 finished
    # 5
    # in Thread-1,wait 01 is triggered
    # in Thread-2,wait 02 finished
    # 6
    # in Thread-2,notify 02 is triggered
    # in Thread-2,wait 02 is triggered
    # in Thread-1,wait 01 finished
    # in Thread-1,notify 01 is triggered
    # in Thread-1,notify 01 finished
    # in Thread-2,wait 02 finished

    # Process finished with exit code 0
# +++++++++++++++++++++++++++++++++++++++++
#如果把tar04和tar05都写成先nofity再wait的形式;
if 1 == 1:
    pass
    # 就会导致在执行最后一步循环时，tar04结束之后没有运行notify函数通知处于等待状态的tar05被唤醒，
    # 从而一直hang在那里，运行结果如下：
    # press Enter to go mod A,else: press other key:  p
    # Thread-1 is started
    # 1
    # in Thread-1,notify 01 is triggered
    # in Thread-1,wait 01 is triggered
    # Thread-2 is started
    # 2
    # in Thread-2,notify 02 is triggered
    # in Thread-2,wait 02 is triggered
    # in Thread-1,wait 01 finished
    # 3
    # in Thread-1,notify 01 is triggered
    # in Thread-1,wait 01 is triggered
    # in Thread-2,wait 02 finished
    # 4
    # in Thread-2,notify 02 is triggered
    # in Thread-2,wait 02 is triggered
    # in Thread-1,wait 01 finished
    # 5
    # in Thread-1,notify 01 is triggered
    # in Thread-1,wait 01 is triggered
    # in Thread-2,wait 02 finished
    # 6
    # in Thread-2,notify 02 is triggered
    # in Thread-2,wait 02 is triggered
    # in Thread-1,wait 01 finished

# +++++++++++++++++++++++++++++++++++++++++
#对于tar04和tar05的情况：假如两个Condition函数运行的总次数是奇数，比如：
if 1 == 1:
    pass
    # tar04执行for循环的次数为3次，for i range(1,6,2) 循环打印的是1，3，5
    # tar05执行for循环的次数为2次，for i range(2,6,2) 循环打印的是2，4
    # 那么tar04写成先wait再notify，tar05写成先notify再wait的话，线程2可以结束，但是线程1无法结束。
    # 因为tar05少了一个wait来通知tar04继续运行notify
    # 程序的运行结果如下：
    # press Enter to go mod A,else: press other key:  p
    # Thread-1 is started
    # 1
    # in Thread-1,wait 01 is triggered
    # Thread-2 is started
    # 2
    # in Thread-2,notify 02 is triggered
    # in Thread-2,wait 02 is riggered
    # in Thread-1,wait 01 finished
    # in Thread-1,notify 04 is tgriggered
    # in Thread-1,notify 04 finished
    # 3
    # in Thread-1,wait 01 is triggered
    # in Thread-2,wait 02 finished
    # 4
    # in Thread-2,notify 02 is triggered
    # in Thread-2,wait 02 is riggered
    # in Thread-1,wait 01 finished
    # in Thread-1,notify 04 is tgriggered
    # in Thread-1,notify 04 finished
    # 5
    # in Thread-1,wait 01 is triggered
    # in Thread-2,wait 02 finished

# ++++++++++++++++++++++
# 无论两个Condition函数执行for循环的次数为奇数还是偶数，都不能写成先同时先notify再wait的形式。
if 1 == 1:
    pass
    # 假如把tar04和tar05都写成先nofity再wait的样子，两个线程依然会交替运行，
    # 但是就会导致在最后一个循环时线程1已经由于notify而结束，但是没有wait来触发tar05继续执行tar05的nofity
    # 所以 tar05会一直hang在那里
    # 结果如下：
    # press Enter to go mod A, else: press other key: p
    # Thread - 1 is started
    # 1
    # in Thread - 1, notify 01 is triggered
    # in Thread - 1, wait 01 is triggered
    # Thread - 2 is started
    # 2
    # in Thread - 2, notify 02 is triggered
    # in Thread - 2, wait 02 is riggered
    # in Thread - 1, wait 01 finished
    # 3
    # in Thread - 1, notify 01 is triggered
    # in Thread - 1, wait 01 is triggered
    # in Thread - 2, wait 02 finished
    # 4
    # in Thread - 2, notify 02 is triggered
    # in Thread - 2, wait 02 is riggered
    # in Thread - 1, wait 01 finished
    # 5
    # in Thread - 1, notify 01 is triggered
    # in Thread - 1, wait 01 is triggered
    # in Thread - 2, wait 02 finished
    # 6
    # in Thread - 2, notify 02 is triggered
    # in Thread - 2, wait 02 is riggered
    # in Thread - 1, wait 01 finished

# +++++++++++++++++++
# 关于Condition
if 1 == 1:
    pass
    # 当你在程序的末尾启动# print('+++++++++++++++finished+++++++++++++++')时，会看到下面的情况。
    # 也就是说，以Condition控制的多线程，不会导致程序在多线程执行完毕之前被阻塞，
    # 所以，当你Condition多线程后面还有代码需要使用Condition多线程的结果时，你要特别留意这个问题。
    # 结果如下
    # Thread-1 is started
    # 1
    # +++++++++++++++finished+++++++++++++++
    # 后面的结果省略